Implicit differentiation

​​In a nutshell

If a function cannot easily be arranged such that you have it in the form $$y=f(x)$$ or $$x=g(y)$$, then you cannot differentiate explicitly, which is what you have been doing so far. Instead, you can differentiate the function as it is implicitly. ​

Explicit vs. implicit

A function of the form $$y=f(x)$$ or $$x=g(y)$$ is given explicitly. On one side is a $$y$$ or an $$x$$, and on the other side is a function in terms of the other variable. A function in the form $$f(x,y)$$ is a function in terms of both $$x$$​ and $$y$$. The equation $$0=f(x,y)$$ is given implicitly. 

Example 1

The following functions are given implicitly.

$$4xy+x^2-y^3=6\\\sin(x-y)=0\\\frac{x}{y^2}-6xy=-9^x$$​​

Using the chain rule

Recall that when you denote differentiation, you use $$\frac{\text d}{\text dx}$$​, meaning to differentiate with respect to $$x$$​. It is commonly seem as $$\frac{\text dy}{\text dx}$$​. meaning you are differentiating $$y$$​ with respect to $$x$$​.

Suppose you have a function of $$y$$, $$f(y)$$, but since it is part of a larger, implicit function, you need to differentiate with respect to $$x$$, you will use the chain rule. In this case, you will use it in this form:

​$$\frac{\text d}{\text dx}\rightarrow\frac{\text dy}{\text dx}\times\frac{\text d}{\text dy}$$​​

In other words, you have converted the differentiation operation into one in terms of $$y$$. All you have to do is to multiply by $$\frac{\text dy}{\text dx}$$. The general form when differentiating a function $$f(y)$$ with respect to $$x$$ is

$$\boxed{\frac{\text d}{\text dx}(f(y))=f'(y)\frac{\text dy}{\text dx}}$$​​

Example 2

Differentiate the following with respect to $$x$$:

$$5y^2+7y=4x^3$$​​

Apply $$\frac{\text d}{\text dx}$$ to both sides:

$$\frac{\text d}{\text dx}(5y^2+7y)=\frac{\text d}{\text dx}(4x^3)$$​​

The right-hand side can be done as expected, since it is in terms of $$x$$:

​$$\frac{\text d}{\text dx}(5y^2+7y)=12x^2$$​​

The left-hand side however requires the chain rule:

$$\begin{aligned}\frac{\text dy}{\text dx}\times\frac{\text d}{\text dy}(5y^2+7y)&=12x^2\\\frac{\text dy}{\text dx}(10y+7)&=12x^2\end{aligned}$$​​

Rearranging this to obtain the equation for the gradient gives:

$$\underline{\frac{\text dy}{\text dx}=\frac{12x^2}{10y+7}}$$​​

Note: When implicitly differentiating, it is common that your expression for $$\frac{\text dy}{\text dx}$$ will be a function of both $$x$$ and $$y$$.

Example 3

Differentiate the following with respect to $$x$$:

$$0=2y+xy^3$$​​

Apply $$\frac{\text d}{\text dx}$$ to both sides:

$$\begin{aligned}\frac{\text d}{\text dx}(0)&=\frac{\text d}{\text dx}(2y+xy^3)\\0&=\frac{\text d}{\text dx}(2y+xy^3)\end{aligned}$$​​

To differentiate the right-hand side, approach term by term:

$$\begin{aligned}0&=\frac{\text d}{\text dx}(2y)+\frac{\text d}{\text dx}(xy^3)\\&=\frac{\text dy}{\text dx}\frac{\text d}{\text dy}(2y)+\frac{\text d}{\text dx}(xy^3)\\&=2\frac{\text dy}{\text dx}+\frac{\text d}{\text dx}(xy^3)\end{aligned}$$​​

To differentiate the second term, you will use the product rule:

$$\begin{aligned}0&=2\frac{\text dy}{\text dx}+\frac{\text d}{\text dx}(xy^3)\\&=2\frac{\text dy}{\text dx}+\left(\frac{\text d}{\text dx}(x)\right)y^3+x\left(\frac{\text d}{\text dx}(y^3)\right)\\&=2\frac{\text dy}{\text dx}+(1)y^3+x\left(\frac{\text dy}{\text dx}\frac{\text d}{\text dy}(y^3)\right)\\&=2\frac{\text dy}{\text dx}+y^3+x\left(\frac{\text dy}{\text dx}(3y^2)\right)\\&=2\frac{\text dy}{\text dx}+y^3+3xy^2\frac{\text dy}{\text dx}\end{aligned}$$​​

You can now rearrange this to give an equation for $$\frac{\text dy}{\text dx}$$:

$$\begin{aligned}0&=2\frac{\text dy}{\text dx}+y^3+3xy^2\frac{\text dy}{\text dx}\\&=\frac{\text dy}{\text dx}(2+3xy^2)+y^3\\\frac{\text dy}{\text dx}&=\underline{-\frac{y^3}{2+3xy^2}}\end{aligned}$$​​

​

Common derivatives

Below are some frequently occuring derivatives. You may want to commit them to memory.

​$$\begin{aligned}\frac{\text d}{\text dx}(y^n)&=ny^{n-1}\frac{\text dy}{\text dx}\\\frac{\text d}{\text dx}(xy)&=x\frac{\text dy}{\text dx}+y\end{aligned}$$​​

where $$n$$ is a constant.